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x → 0 lim ​ x tan x cos x − cos 2 x ​ = ...

                    

  1.  2    

  2. 1 comma 5        

  3. 1  

  4. 2 over 3             

  5. 0 comma 5  

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P. Anggrayni

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

jawaban yang benar adalah B.

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Pembahasan

Ingatlah rumus selisih cosinus serta sifat-sifat limit fungsi Dengan menggunakan sifat-sifat limit fungsi danrumus selisih cosinus tersebut, persoalan limit di atas dapat diselesaikan dengan cara sebagai berikut: Maka, . Oleh karena itu, jawaban yang benar adalah B.

Ingatlah rumus selisih cosinus

cos space straight A minus cos space straight B equals negative 2 space sin space 1 half open parentheses straight A plus straight B close parentheses space sin space 1 half open parentheses straight A minus straight B close parentheses

serta sifat-sifat limit fungsi

  • limit as x rightwards arrow c of k times f left parenthesis x right parenthesis equals k times limit as x rightwards arrow c of f left parenthesis x right parenthesis
     
  • limit as x rightwards arrow c of open square brackets f left parenthesis x right parenthesis times g left parenthesis x right parenthesis close square brackets equals limit as x rightwards arrow c of f left parenthesis x right parenthesis times limit as x rightwards arrow c of g left parenthesis x right parenthesis
     
  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator b x end fraction equals a over b
     
  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator tan space b x end fraction equals a over b

Dengan menggunakan sifat-sifat limit fungsi dan rumus selisih cosinus tersebut, persoalan limit di atas dapat diselesaikan dengan cara sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 0 of space fraction numerator cos space x minus cos space 2 x over denominator x space tan space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative open parentheses cos space 2 x minus cos space x close parentheses over denominator x space tan space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative open parentheses negative 2 space sin 1 half open parentheses 2 x plus x close parentheses space sin 1 half open parentheses 2 x minus x close parentheses close parentheses over denominator x space tan space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator 2 space sin space begin display style 3 over 2 end style x space sin space 1 half x over denominator x space tan space x end fraction end cell row blank equals cell 2 times limit as x rightwards arrow 0 of space open parentheses fraction numerator sin space begin display style 3 over 2 end style x over denominator x end fraction times fraction numerator sin space begin display style 1 half end style x over denominator tan space x end fraction close parentheses end cell row blank equals cell 2 times limit as x rightwards arrow 0 of space fraction numerator sin space begin display style 3 over 2 end style x over denominator x end fraction times limit as x rightwards arrow 0 of fraction numerator sin space begin display style 1 half end style x over denominator tan space x end fraction end cell row blank equals cell 2 times open parentheses fraction numerator begin display style 3 over 2 end style over denominator 1 end fraction close parentheses times open parentheses fraction numerator begin display style 1 half end style over denominator 1 end fraction close parentheses end cell row blank equals cell 3 over 2 end cell row blank equals cell 1 comma 5 end cell end table  

Maka, limit as x rightwards arrow 0 of space fraction numerator cos space x minus cos space 2 x over denominator x space tan space x end fraction equals 1 comma 5

Oleh karena itu, jawaban yang benar adalah B.

Latihan Bab

Konsep Kilat

Limit Fungsi Trigonometri

Limit Tak Hingga

Kekontinuan dan Asimtot

190

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